Table of contents  
Convolution Theorem  
Transfer Functions for Linear Systems  
The Concept of Pole & Zero  
Transfer Function of Closed Loop System 
The Transfer Function (TF) of a System is the ratio of the output to the input of a system, in the Laplace domain considering its initial conditions and equilibrium point to be zero.
If we have an input function of X(s), and an output function Y(s), we deﬁne the transfer function H(s).
H(s) = Y(s)/X(s)
Since we know that in the time domain, generally, we define the input to a system as x(t), and the output of the system as y(t). The relationship between the input and the output is represented as the impulse response, h(t).
We can use the following equation to deﬁne the impulse response:
h(t) = y(t) / x(t)
Convolution in Time Domain
If we have the system input and the impulse response of the system, we can calculate the system output using the convolution operation as such as
y(t) = h(t) ∗ x(t).
TimeInvariant System Response
Example: Impulse Response
Consider a linear input/output system described by the controlled differential equation
where u is the input and y are the output.
To investigate how a linear system responds to the exponential input u(t) = e^{s}^{t} we consider the state space system
dx/dt = Ax + Bu
y = Cx + Du
Let the input signal be u(t) = e^{st} and assume that s ≠ λ_{i}(A), where i = 1,...,n, where λ_{i}(A) is the i^{th }eigenvalue of A. then
Since s ≠ λ(A) the integral can be evaluated and we get
& Output y(t) = Cx(t) + Du(t)
= Ce^{At}{x(0) − (sI − A)^{−1}B} + [D + C(sI − A)^{−1}B]e^{st}
Note: One term of the output is proportional to the input u(t) = e^{st}. This term is called the pure exponential response.
Eﬀects of Poles and Zeros
Note: If the Poles or Zero lie on the imaginary axis then it must be simple (order only 1) then the system is said to be Marginally Stable, If the Order of multiplicity of Poles or Zeros on Imaginary Axis is more than 1 in that case system will become Unstable.
Block diagram of Closed Loop SystemE(s) = Actuating or Error Signal
X(s) = Reference Input Signal.
G(s) = Forward Path Transfer Function.
Y(s) = Output Signal.
H(s) = Feedback Transfer Function.
B(s) = Feedback Signal.
So, the transfer function of the closed loop system is Y(s)/X(s).
From the block diagram,
Y(s) = G(s).E(s) .........1
B(s) = H(s).Y(s) .........2
E(s) = X(s) + B(s) .........3a (For positive feedback)
= X(s)  B(s) ..........3b (For negative feedback)
Y(s) = G(s).[X(s)B(s)]
Y(s) = G(s).X(s)  G(s).B(s) .........4
Put the value of B(s) from eq.2 in eq.4
Y(s) = G(s).X(s)  G(s).H(s).Y(s)
Y(s) + G(s).H(s).Y(s) = G(s).X(s)
Y(s){1 + G(s).H(s)} = G(s).X(s)
Y(s)/X(s) = G(s) / {1 +G(s).H(s)} .........5
For positive feedback system we will use eq.3a and repeat the all the same steps and we will get the transfer function as:
Y(s)/X(s) = G(s) / {1  G(s).H(s)} .........6
For unity feedback H(s) = 1, the eq.5 & eq.6 will become
Y(s)/X(s) = G(s)/{1 + G(s)} .......for negative unity feedback
Y(s)/X(s) = G(s)/{1  G(s)} ........for positive unity feedback
54 docs46 tests

1. What is the Convolution Theorem in electrical engineering? 
2. How can the Transfer Function be used to analyze linear systems? 
3. What is the significance of poles and zeros in a Transfer Function? 
4. How does the Transfer Function of a closedloop system differ from an openloop system? 
5. How is the Transfer Function used in electrical engineering? 
54 docs46 tests


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